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8. Solve for x:(i) 1+6++ 16+..-+x=148 |
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Answer» Given 1 + 6+ 11+ 16 + ............+x = 148take the AP :1,6,11,16,..........,xIn this APa = 1d = 6-1 = 5Given Sn = 148we know that Sn = n/2[2a+ (n-1)d]⇒n/2 [ 2a +(n-1)d] = 148⇒n[2(1) + (n-1)5 ] = 148×2⇒n[ 2 +5n - 5 ] = 296⇒n [ -3 + 5n ] = 296⇒-3n + 5n² = 296⇒5n² - 3n -296 = 0⇒ 5n² - 40n + 37n - 296 = 0⇒5n( n - 8) + 37( n - 8) = 0⇒(n - 8) (5n + 37) = 0⇒n-8 = 0 or 5n +37 = 0⇒n = 8 or n= -37/5. As n is the no.of terms in the AP can not be fractional and negative∴n=8Theirfore x is the 8th termTn = T8 = a + (n-1)d = 1 + (8-1)5 = 1+7(5) = 1+35 = 36∴8th term = x = 36. eighth term is thirty six (36) |
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